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Conical pendulum in rotating frame

  1. Feb 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A pendulum of length [itex] l [/itex] at the north pole is moving in a circle to the east at an angle [itex] \theta [/itex] to the vertical. It has some period [itex] T_E [/itex] as measured in the rotating Earth frame. The experiment is then repeated except now the pendulum is moving to the west with period [itex] T_W [/itex]. The question asks which period is longer, and to calculate the relative time difference in the periods.

    2. Relevant equations

    In the rotating frame, the equation of motion of the pendulum involves the Coriolis force and a modified gravitational acceleration due to the rotation of the Earth: [tex] m\mathbf{a'} = m\mathbf{g'} + \mathbf{T} -2m(\mathbf{\omega '} \times \mathbf{v'}) [/tex]

    where T is the tension in the wire.

    3. The attempt at a solution
    Since the Earth rotates counterclockwise as viewed looking down on the north pole, the pendulum that moves to the east has a longer period because the Earth's rotation means it has to "catch up" with the rotating ground (I think?). However, I am at a loss as to how to calculate the period from equation of motion.
     
  2. jcsd
  3. Feb 12, 2017 #2

    TSny

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    Can you describe in more detail the "modified gravitational acceleration" ##\mathbf{g'}##? What is its magnitude and direction?

    What does the symbol ##\mathbf{\omega '}## stand for?

    What about the "centrifugal force" in the rotating frame? Or does your ##\mathbf{g'}## include that?

    Yes, the "catching up" is from the point of view of an observer in a nonrotating. In fact, you can work the problem by analyzing it from the nonrotating frame. But, if you want to work it out in the rotating frame, then follow the usual recipe of starting with a free body diagram, etc.
     
    Last edited: Feb 12, 2017
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